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Let $\{ v_1, v_2, v_3, \dotsc, v_n \}$ be an orthonormal basis of $V$. Show that for any vectors $w$ and $z$ of $V$: $$ \langle w,z \rangle = \sum_{k=1}^{n} \langle w,v_k \rangle \langle v_k,z \rangle. $$

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  • $\begingroup$ Hint: try to prove that $w = \sum_{i = 1}^n \langle w, v_i\rangle \cdot v_i $. $\endgroup$ – Xiangxiang Xu Sep 2 '18 at 9:09
  • $\begingroup$ @XiangxiangXu $w=\sum_{k=1}^{n}a_kv_k$. If we take inner product with $v_k$ on both sides, we get, $a_k=\langle w,v_k \rangle$ $\endgroup$ – Christy Sep 2 '18 at 9:20
  • $\begingroup$ Then suppose $w = \sum_{i = 1}^n a_i v_i, z = \sum_{i = 1}^n b_i v_i$, can you show that $\langle w, z \rangle = \sum_{i = 1}^n a_i b_i$? $\endgroup$ – Xiangxiang Xu Sep 2 '18 at 9:32
  • $\begingroup$ @Chintu Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE $\endgroup$ – user Oct 23 '18 at 20:43
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We have that

  • $w=\sum_{i=1}^{n} a_i v_i$
  • $z=\sum_{j=1}^{n} b_j v_j$

then

$$\langle w,z \rangle=\langle\sum_{i=1}^{n} a_i v_i,\sum_{j=1}^{n} b_j v_j \rangle=\sum_{k=1}^{n} a_kb_k=\sum_{k=1}^{n} a_kv_k^T\cdot v_kb_k= \sum_{k=1}^{n} \langle w,v_k\rangle\langle v_k,z\rangle$$

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  • $\begingroup$ you have used dot product definition. can we prove using only the generalised inner product definition? $\endgroup$ – Christy Sep 2 '18 at 9:23
  • $\begingroup$ @mathamity Yes the main steps are equivalent, that is $\langle v_i,v_j \rangle=0$ and $\langle w,v_k \rangle=a_k$. $\endgroup$ – user Sep 2 '18 at 9:25
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Since $\{v_1, \ldots, v_n\}$ is an orthonormal basis, we have $\langle v_k, v_j\rangle = \delta_{kj}$ and $x = \sum_{i=1}^n \langle x, v_i\rangle v_i$ for all $x \in V$.

Therefore $$\langle w,z \rangle= \left\langle \sum_{k=1}^n \langle w, v_k\rangle v_k, \sum_{j=1}^n \langle z, v_j\rangle v_k\right\rangle = \sum_{k=1}^n\sum_{j=1}^n \langle w,v_k\rangle\langle v_j,z\rangle \underbrace{\langle v_k, v_j\rangle}_{\delta_{kj}} = \sum_{k=1}^{n} \langle w,v_k\rangle\langle v_k,z\rangle$$

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